Unformatted text preview: ty. 4.6 Lyapunov Functions
A continuous function
is a strong Lyapunov function for an equilibrium of a flow
if there exists an open neighborhood of such that
is a weak Lyapunov function if the same holds except
. for all is positive definite on a neighborhood of
strong Lyapunov function is positive definite in some neighborhood of an equilibrium point .
Theorem 4.7 (Lyapunov Functions) Let be an equilibrium point of a flow
. If is a
weak Lyapunov function in some neighborhood of , then is stable. If is a strong
Lyapunov function, then is asymptotically stable.
Proof. First prove that
and is stable if is a weak Lyapunov function. Choose
so small that
. Choose so small that
. Then if
is stable. Next prove that is asymptotically stable if is a strong Lyapunov function. Note that is
still stable as argued above. Consider a trajectory whose initial point is any
. We will
show that is the only limit point of the flow. Together with the fact that is stable, this
Chapter 4A Finally prove that is the only limit point of the flow. Consider any sequence of times
has a convergent subsequence. For simplicity, denote the subsequence of times
. Because is a decreasing function
However, because is not an equilibrium point,
continuity of , for for any . Fix large enough, we must also have , By . But then for , which contradicts the chain of inequalities displayed above. □
Example. Show that the origin is an asymptotically stable state for the system
For a candidate Lyapunov function, try a very simple positive definite function with minimum at
the origin. Let
. Then by the chain rule
for . is a strong Lyapunov function. The origin is an asymptotically stable equilibrium. ■
Recall Hamiltonian systems. Let
and and consider
. often represents the energy of a system. It is constant along trajectories:
If is an equilibrium, then Is zero at the equilibrium and constant along trajectories. If is positive definite in some
neighborhood of the equilibrium, then it is a weak Lyapunov function.
Example. The simple pendulum. The Hamiltonian is
, , leading to 13
Chapter 4A .
sketch support, rod, bob, angle is an equilibrium point. . Consider
. is positive definite in a neighborhood of the origin, so
is a stable equilibrium point. is a weak Lyapunov function. sketch phase portrait near origin. Nearly circular orbits surround the origin.
The linearized system is a center. ■
Example. The damped pendulum. The pendulum experiences friction proportional to velocity.
We now expect the system to lose energy. The system is no longer Hamiltonian.
an equilibrium point. Is
still a Lyapunov function? is still .
, is still a weak Lyap nov f n tion. Lyap n...
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This document was uploaded on 02/24/2014 for the course MATH 512 at Washington State University .
- Fall '14